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The integral 01/2ex2x2(1x)3/2(1+x)1/2dx is equal to

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a
3e
b
3e−1
c
e3
d
e3−1

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detailed solution

Correct option is B

Let I=∫01/2 ex2−x2(1−x)3/2(1+x)1/2dx=∫01/2 ex1−x2+1(1−x)1-x2dx=∫01/2 ex1−x21−x+1(1−x)1−x2dxAs ddx1−x21−x=(1−x)−x1−x2+1−x2(1−x)2=−x+x2+1−x2(1−x)21−x2=1(1−x)1−x2∴I=ex1−x21−x01/2=e3/42/2−1=3e−1


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